Question: Complete the square to solve for $x$. $4x^{2}-8x+3 = 0$
Solution: First, divide the polynomial by $4$ , the coefficient of the $x^2$ term. $x^2 - 2x + \dfrac{3}{4} = 0$ Move the constant term to the right side of the equation. $x^2 - 2x = -\dfrac{3}{4}$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $-2$ , so half of it would be $-1$ , and squaring it gives us ${1}$ $x^2 - 2x { + 1} = -\dfrac{3}{4} { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x - 1 )^2 = \dfrac{1}{4}$ Take the square root of both sides. $x - 1 = \pm\dfrac{1}{2}$ Isolate $x$ to find the solution(s). $x = 1\pm\dfrac{1}{2}$ The solutions are: $x = \dfrac{3}{2} \text{ or } x = \dfrac{1}{2}$ We already found the completed square: $( x - 1 )^2 = \dfrac{1}{4}$